Answer

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**Hint:**From Kepler’s law, the square of the time period of revolution of a celestial body is directly proportional to the cube of the radius of the circular path that it travels on. Use this theory to form an equation for the time period and find the relation between the two time periods of the moon.

**Formula used:**${{T}^{2}}=k{{r}^{3}}$

**Complete step by step answer:**

When a body revolves around another body, we define its period of revolution. The period of revolution is the time taken by the revolving body to complete one full revolution.

Here, in the given case, the moon is revolving around the earth. Let us assume that the moon revolves around the earth as a circular path.

From Kepler’s law, the square of the time period of revolution of a celestial body is directly proportional to the cube of the radius of the circular path that it travels on.

i.e. ${{T}^{2}}\propto {{r}^{3}}$

$\Rightarrow {{T}^{2}}=k{{r}^{3}}$ ….. (i), where k is a proportionality constant.

Let the radius of the circular path of the moon be r and its period of revolution be T.

Then, we know that ${{T}^{2}}=k{{r}^{3}}$…. (ii)

It is said that the radius of the moon’s orbit is doubled. Let the new time period of the moon be T’.

Substitute r = 2r and T = T’ is equation (i).

$\Rightarrow T{{'}^{2}}=k{{(2r)}^{3}}$

$\Rightarrow T{{'}^{2}}=8k{{r}^{3}}$ …. (iii).

Now divide (iii) by (ii).

$\Rightarrow \dfrac{T{{'}^{2}}}{{{T}^{2}}}=\dfrac{8k{{r}^{3}}}{k{{r}^{3}}}$

$\Rightarrow T{{'}^{2}}=8{{T}^{2}}$

$\Rightarrow T'=\sqrt{8}T={{2}^{3/2}}T$.

Therefore, when the radius of the moon’s orbit is doubled, its period becomes ${{2}^{3/2}}$ times its initial period.

**So, the correct answer is “Option D”.**

**Note:**We know that the time period of revolution of any celestial body depends on the radius of the circle if the body is moving in a circular orbit.

When the celestial body is moving in an elliptical orbit, we consider the length of its semi-major axis.

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